- Asked 4/31
- 2080 Baisakh · 4 marks
Area by DMD, traverse ABCDEA (-5.693)
Short answer
Using the Double Meridian Distance method with the given latitudes and departures, area = 2028.41 m² (0.2028 ha).
Calculate the area of the following traverse ABCDEA by double meridian distance method.
| Line | AB | BC | CD | DE | EA |
|---|---|---|---|---|---|
| Latitude (m) | -5.693 | -21.361 | -28.201 | +1.103 | +54.242 |
| Departure (m) | -33.990 | -13.911 | +18.867 | +28.608 | +0.426 |
The latitudes sum to 0.090 m and the departures to 0.000 m. The latitude misclosure of 0.090 m is small (about 1 in 2,000 of the perimeter of about 177 m), so the given figures are used without adjustment.
DMD method
DMD rule (taking the meridian through the starting station):
-
DMD of the first line = departure of the first line.
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DMD of any line = DMD of the previous line + departure of the previous line + departure of the line itself.
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DMD of the last line = departure of the last line with opposite sign (a check).
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Double area = Σ (latitude × DMD), and area = ½ |Σ L × DMD|.
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DMD(AB) = D(AB) = -33.990
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DMD(BC) = -33.990 + (-33.990) + (-13.911) = -81.891
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DMD(CD) = -81.891 + (-13.911) + (18.867) = -76.935
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DMD(DE) = -76.935 + (18.867) + (28.608) = -29.460
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DMD(EA) = -29.460 + (28.608) + (0.426) = -0.426
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Check: DMD(EA) = -0.426 = −D(EA) = -0.426, so the DMDs are correct.
| Line | Latitude L (m) | Departure D (m) | DMD (m) | L × DMD (m²) |
|---|---|---|---|---|
| AB | -5.693 | -33.990 | -33.990 | 193.505 |
| BC | -21.361 | -13.911 | -81.891 | 1749.274 |
| CD | -28.201 | 18.867 | -76.935 | 2169.644 |
| DE | 1.103 | 28.608 | -29.460 | -32.494 |
| EA | 54.242 | 0.426 | -0.426 | -23.107 |
| Sum | 0.090 | 0.000 | 4056.821 |
Answer: Area = 2028.41 m² (0.2028 ha).
(If the latitudes are first balanced by Bowditch's rule, the area changes only slightly, to about 2030 m².)
✓ Answer checked by Claude Opus 5.5 (high effort)
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